Question

a train is moving at a speed of 50m/s and reduces it's speed to 10m/s by covering a distance of 240m then find retardation of the train ​

Answer 1

Given :-

• A train is moving at a speed of 50m/s and reduces it's speed to 10m/s by covering a distance of 240m.

To find :-

• Retardation of the train.

Solution :-

• Initial velocity (u) = 50m/s

• Final velocity (v) = 10m/s

• Distance (s) = 240 m

According to third equation of motion

→ v² = u² + 2as

Where " v " is final velocity, " u " is initial velocity, " a " is acceleration and " s " is distance covered by object.

• According to the question

→ v² = u² + 2as

→ (10)² = (50)² + 2 × a × 240

→ 100 = 2500 + 480a

→ 100 - 2500 = 480a

→ -2400 = 480a

→ a = -2400/480

→ a = - 5m/s

• Negative sign shows retardation.

Hence,

• Retardation of train is 5 m/s

More to know :-

• v = u + at (First equation of motion)

• s = ut + ½ at² (second equation of motion)

Answer 2

Answer:

Question

a train is moving at a speed of 50m/s and reduces it's speed to 10m/s by covering a distance of 240m then find retardation of the train

Given

$\small \: {initial \: velocity \: = 50m/s}$

$final \: velocity \: = 10 \: \: m \: in \: s$

$distance \: = 240 \: m$

Solution

$\longrightarrow \: {v}^{2} = {u}^{2} + 2as$

$\longrightarrow \: {10}^{2} = {20}^{2} + 2 × a × 240$

$\longrightarrow \: 100 = 2500 + 480a$

$\longrightarrow \: 100-2500 = 480a$

$\longrightarrow \: -2400 = 480a$

$\longrightarrow \: a = \frac {-2400}{480}$

$\longrightarrow \: a = -5m / s$